CodeForces 113B Petr#

题目链接：http://codeforces.com/problemset/problem/113/B

题目大意：

多组数据
每组给定3个字符串T，Sbeg，Sed，求字符串T中有多少子串是以Sbeg开头，Sed结尾的

分析：

　　难点在哈希函数的编写，如果直接存string会爆内存，不能用STL自带哈希函数，用了会Wa。

代码如下：

 #include <bits/stdc++.h>
 using namespace std;

 #define rep(i,n) for (int i = 0; i < (n); ++i)
 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)

 #define pr(x) cout << #x << " = " << x << "  "
 #define prln(x) cout << #x << " = " << x << endl

 #define ALL(x) x.begin(),x.end()
 #define INS(x) inserter(x,x.begin())

 #define ms0(a) memset(a,0,sizeof(a))
 #define msI(a) memset(a,inf,sizeof(a))

 #define pii pair<int,int>
 #define piii pair<pair<int,int>,int>
 #define mp make_pair
 #define pb push_back
 #define fi first
 #define se second

 inline int gc(){
     static const int BUF = 1e7;
     static char buf[BUF], *bg = buf + BUF, *ed = bg;

     if(bg == ed) fread(bg = buf, , BUF, stdin);
     return *bg++;
 } 

 inline int ri(){
     int x = , f = , c = gc();
     for(; c<||c>; f = c=='-'?-:f, c=gc());
     for(; c>&&c<; x = x* + c - , c=gc());
     return x*f;
 }

 typedef long long LL;
 typedef unsigned long long uLL;
 const LL mod = 1e9 + ;
 const int maxN =  + ;

 string T, Sbeg, Sed;
 unordered_set< LL > sll;
 int beg[maxN], begLen; // 记录 Sbeg出现的位置
 int ed[maxN], edLen; // 记录 Sed出现的位置 

 // h为T的后缀哈希数组
 // h[i]表示T从i位置开始的后缀的哈希值
 // h[i] = T[i] + T[i+1]*key + T[i+2]*key^2 + ……+ T[i+len-1-i]*key^len-1-i
 // xp为基数数组
 // xp[i] = key^i
 LL xp[maxN], h[maxN];
 const LL key = 1e9 + ;

 // 求起点为s，长为len的子串的哈希值
 // Hash(i, len) = T[i] + T[i+1]*key + T[i+2]*key^2 + ……+ T[i+len-1]*key^len-1
 LL Hash(int s, int len) {
     return h[s] - h[s + len] * xp[len];
 }

 void HashInit(const char* s, LL* h, int len) {
     xp[] = ;
     For(i, , maxN - ) xp[i] = xp[i - ] * key;

     h[len] = ;
     rFor(i, len - , ) h[i] = h[i + ] * key + s[i];
 }

 int main(){
     while(cin >> T >> Sbeg >> Sed) {
         HashInit(T.c_str(), h, (int)T.size());

         int ans = ;
         sll.clear();
         begLen = edLen = ;

         int p = ;
         while(p < T.size()) {
             int t = T.find(Sbeg.c_str(), p);
             if(t == string::npos) break;
             beg[begLen++] = t;
             p = t + ;
         }

         p = ;
         while(p < T.size()) {
             int t = T.find(Sed.c_str(), p);
             if(t == string::npos) break;
             ed[edLen++] = t;
             p = t + ;
         }

         int i = , j = ;
         while(i < begLen && j < edLen) {
             if(beg[i] <= ed[j]) {
                 For(k, j, edLen - ) {
                     if(beg[i] + Sbeg.size() > ed[k] + Sed.size()) continue;
                     sll.insert(Hash(beg[i], ed[k] + Sed.size() - beg[i]));
                 }
                 ++i;
             }
             else ++j;
         }

         cout << sll.size() << endl;
     }
     return ;
 }